Lecture11: Complexnumbersfordiﬀerentialequations October14 ... · Lecture11: Complexnumbersfordiﬀerentialequations October14,2010 Announcements I Homework5dueOctober19. 10%penaltyifmorethan5minuteslate

Lecture 11: Complex numbers for differential equations

October 14, 2010Announcements

I Homework 5 due October 19. 10% penalty if more than 5 minutes late.I Homework 5 revised: Bonus problem.I Next office hours: Monday, 2–3PM, Math 614.


I More complex arithmetic

Complex numbers in calculus

Differential equations: Intro

Solving second-order linear differential equations

Complex division

We can also divide complex numbers:1

a + bi=

1a + bi

a − bia − bi

=a − bi

a2 − b2i2

=a − bia2 + b2 .

What is this geometrically?The number a − bi is called the complex conjugate of a + bi , written a + bi .Always have z z = ‖z‖2.

Complex numbers are unique: IC is R plus a square root of −1. Why not add some more elements?

Fundamental Theorem of AlgebraAny non-constant polynomial p(z) has a root in C, even if we allow complexcoefficients.

This is very important, but a bit beyond us. The first (essentially) correct proof wasdue to Gauss in 1799, after more than 50 years of attempts. He gave at least four.Why did the Greatest Mathematician give four proofs? Because it’s such a remarkabletheorem. We added artificially a solution to x2 + 1 = 0, and suddenly we have solutionsto everything!

ExamplesI Find a solution to z2 = i .I Find a solution to z6 = −1.I Find all solutions to z6 = 1.

Complex numbers are unique: IICan we find other nice products, besides those we already know about?

DefinitionA vector product is a function that takes two vectors in Rn and gives another (insymbols: Rn × Rn → Rn) that is

I Distributive over addition: ~u × (~v + ~w),I Associative with scalar mult.: (a~v)× ~w = a(~v × ~w).

It has division if, for fixed ~v , ~w 6= 0, we can always solve ~x × ~v = ~w for ~x .

1, 2, 4, 8 Theorem [Frobenius,Hurwitz]The only associative, commutative vector products with division are R and C.Drop commutative: Also have quaternions H (on R4).Also drop associative: Still another: octonions O (on R8).Quaternions are related to the cross product.(Actually, need to assume “normed” in last case.)


More complex arithmetic

I Complex numbers in calculus



Parametric curves in C

Some parametric curves are convenient to express using complex numbers.

ExampleThe curve z(t) = 2e it is the circle around the origin of radius 2.What’s the curve z(t) = 2e it + 1+ 3i?

ExampleThe curve z(t) = (1+ it)2 is (1− t2, 2t), a parabola.

Unifying formulas

You know how to integrate 1/(x2 − 1):∫1

x2 − 1dx =

∫ (1/2x − 1

− 1/2x + 1

)dx = 1

2 ln(x − 1)− 12 ln(x + 1) + C .

You also know how to integrate 1/(x2 + 1):∫1

x2 + 1dx = arctan(x) + C

Complex numbers let you do these two problems in the same way:1

x2 + 1=

i/2x + i

− i/2x − i∫

1x2 − 1

dx =

∫ (i/2x + i

− i/2x − i

)dx = i

2 ln(x + i)− 12 ln(x − i) + C .

These are actually the same, for correct generalization of ln.




I Differential equations: Intro


Idea of differential equations

A differential equation is an equation for a function f (x), involving f and its derivatives.

ExamplesI f ′(x) = f (x). Solution: f (x) = ex .I f ′′(x) = −f (x). Solutions:

f (x) = cos(x), f (x) = sin(x).

f (x) = e ix , f (x) = e−ix .

I f ′′′(x) = x . Solution:

f (x) = x4/24.

Differential equations come up all the time!



ExamplesI f ′(x) = f (x). Solution: f (x) = ex .I f ′′(x) = −f (x). Solutions: f (x) = cos(x), f (x) = sin(x).

f (x) = e ix , f (x) = e−ix .


f (x) = x4/24.





f (x) = e ix , f (x) = e−ix .


f (x) = x4/24.





f (x) = e ix , f (x) = e−ix .

I f ′′′(x) = x . Solution: f (x) = x4/24.


Example: Pendulum

QuestionSuppose you have a pendulum of length l and mass m. What does its motion look like?What is angle θ as a function of t?

AnswerPendulum weight experiences a force gm downwards.Tangential component: gm sin(θ), pointing inwards. Thus:

mlθ′′(t) = −gm sin(θ(t))θ′′(t) = −g

l sin(θ(t))≈ −g

l θ(t).(Why lθ′′? That’s what the acceleration is. Also, needed for units.)Solution (adopted from previous):

θ(t) = A sin(√g

l t)+ B cos

(√gl t).

Example: Pendulum






θ(t) = A sin(√g

l t)+ B cos

(√gl t).

Example: Pendulum






θ(t) = A sin(√g

l t)+ B cos

(√gl t).

Example: Pendulum, cont.Pendulum’s motion is

θ(t) = A cos(√g

l t)+ B sin

(√gl t).

QuestionSuppose I pull the pendulum to an angle of .1 radian and let it go. Describe the motion.

Answer

0.1 cos(√

g/l t).

QuestionHow long do the pendulum’s oscillations take?

Answer

2π√

l/g seconds. (Check: units.)

Independent of size of swing! (As long as oscillations are small.)Observed by Galileo; key to pendulum clocks.


θ(t) = A cos(√g

l t)+ B sin

(√gl t).


Answer0.1 cos

(√g/l t

).


Answer

2π√

l/g seconds. (Check: units.)

Independent of size of swing! (As long as oscillations are small.)Observed by Galileo; key to pendulum clocks.


θ(t) = A cos(√g

l t)+ B sin

(√gl t).


Answer0.1 cos

(√g/l t

).


Answer2π√

l/g seconds. (Check: units.)Independent of size of swing! (As long as oscillations are small.)Observed by Galileo; key to pendulum clocks.

Second-order linear differential equations

These types of equations come up often.

DefinitionA second-order linear differential equation is an equation of the form

af ′′(x) + bf ′(x) + cf (x) = 0.(Second order: only second derivatives appear.)

TheoremA second-order linear differential equation has two basic solutions. Any other solutionscan be written as a linear combination of these two.(Linear combination of f1(x) and f2(x): Af1(x) + Bf2(x), for constants A and B .)Idea: Position and first derivative determine the entire Taylor series.

But wait! What about. . .

We found four different solutions to f ′′(t) = −f (t):cos(t), sin(t),

e it , e−it .

What’s happening?In fact, any two of these solutions can be taken as the basic solutions:

e it = cos(t) + i sin(t) e−it = cos(t)− i sin(t)

cos(t) =e it + e−it

2sin(t) =

e it − e−it

2i.

Basic solutions are not unique.

But wait! What about. . .

We found four different solutions to f ′′(t) = −f (t):cos(t), sin(t),

e it , e−it .

What’s happening?In fact, any two of these solutions can be taken as the basic solutions:

e it = cos(t) + i sin(t) e−it = cos(t)− i sin(t)

cos(t) =e it + e−it

2sin(t) =

e it − e−it

2i.

Basic solutions are not unique.





I Solving second-order linear differential equations

An exampleQuestionWhat are the solutions to f ′′(t) + 2f ′(t) + 5f (t) = 0?

AnswerGuess that the answer might be Aeαt . Get

α2Aeαt + 2αAeαt + 5Aeαt = 0

α2 + 2α+ 5 = 0

α =−2±

√4− 20

2= −1± 2i .

Two basic solutions:f1(t) = e(−1+2i)t = e−te2it f2(t) = e(−1−2i)t = e−te−2it

Alternatively:g1(t) = e−t cos(2t) g2(t) = e−t sin(2t)

Exercise: Solve f ′′(t)− 2f ′(t) + 2f (t) = 0 with f (0) = 1, f ′(0) = 2.




α2 + 2α+ 5 = 0

α =−2±

√4− 20

2= −1± 2i .







α2 + 2α+ 5 = 0

α =−2±

√4− 20

2= −1± 2i .




Linear second-order equations: General caseIn general, to solve af ′′(x) + bf ′(x) + cf (x) = 0:

I Suppose f (x) = eαx .I Get characteristic equation: aα2 + bα+ c = 0.

I Solve for α: get α =−b ±

√b2 − 4ac2

. Call solutions α1 and α2.

I Basic solutions are (usually) eα1x and eα2x ; general solution is Aeα1x + Beα2x .I Find A and B given whatever else you know.

Behavior depends on b2 − 4ac :I b2 − 4ac > 0: Now α1, α2 are real, distinct. Two different growths.I b2 − 4ac < 0: Take a square root of a negative number; two complex roots.

Alternate basic solutions using sin and cos.I b2 − 4ac = 0: Repeated root! Need another basic solution.

Turns out to be xeαx , where α is root of characteristic equation.

Physical interpretation: DampingIn our earlier pendulum example (and also for projectiles), ignored friction/air resistance.(Our pendulum kept swinging forever. . . )

Typical air resistance: Force opposite to velocity and proportional to it: ~Ffric = −b~v .Consider a mass on a spring. The spring has a spring constant k , and ~Fspring = −kr(t).(Here r(t) is the displacement from “rest”.) Now get

mr ′′(t) = −br ′(t)− kr(t)mr ′′ + br ′ + kr = 0.

Cases:I b = 0: undamped: two periodic solutions, no decay.I b > 0, b2 − 4km < 0: underdamped: periodic solutions with an exponential decay.I b2 − 4km = 0: critically damped.I b2 − 4km > 0: over damped: Two exponentially decaying solutions.

Critical damping turns out to decay the fastest. Useful in shock absorbers.

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Lecture11: Complexnumbersfordiﬀerentialequations October14 ... · Lecture11: Complexnumbersfordiﬀerentialequations October14,2010 Announcements I Homework5dueOctober19. 10%penaltyifmorethan5minuteslate